### Table 1 The result fl)r Arabic is impractical for small computers.

### Table 1. Eigenvalues of the Schur complement, for streamline-upwind discretization. We test the preconditioners here with two Krylov subspace methods for solving nonsym- metric systems: the generalized minimum residual method (GMRES) [15], and a simple im- plementation of the quasi-minimum residual method (QMR) [8] based on coupled two-term recurrences without look-ahead. GMRES demonstrates the performance of the preconditioners with the optimal (with respect to the residual norm) Krylov subspace solver. This method is impractical for large problems because its work and storage requirements grow with the iteration count; QMR is a non-optimal alternative that avoids this di culty. Some additional experi- ments with restarted GMRES are presented in Section 4. In all cases we use right-oriented preconditioning, and our convergence criterion is a reduction of 10?6 in the l2{norm of the residual. The action of F?1 and F?t is computed using the LU-factorization in MATLAB. We start from a zero initial guess. Using random initial guesses gave comparable iteration counts 10

"... In PAGE 12: ... We rst consider the bounds of Theorem 1. Table1 shows the extreme real parts and max- imum imaginary parts of the generalized eigenvalues (2.7) of the Schur complement operator, for = 1=10 and 1=100 with the streamline-upwind discretization, on three meshes.... In PAGE 12: ... The analysis also shows that the real parts and largest imaginary parts of the eigenvalues are bounded independently of ; the bound for the smallest real part is proportional to 2. The data of Table1 are in agreement with the upper bounds. Figure 5 plots the smallest real parts on a logarithmic scale, for the streamline-upwind discretization on a 64 64 grid and = 1=20, 1=40, 1=80, and 1=160.... ..."

### Table 4 : Computational Requirements

"... In PAGE 5: ... 43 Table 3: Determining Silhouettes. 47 Table4 : Computational Requirements 54 Table 5: Display Rate Requirements 55 Table 6: Comparison of Complex Surface Methods 56 ... ..."

### Table 3: Computational Requirements

"... In PAGE 15: ...4.2 Computational Requirements To give an idea of the computational requirements of CLNB and CLNN, four measurements are used: (1) running time for training (in second); (2) running time for testing (in second); (3) number of contextual rules generated before pruning; and (4) number of contextual rules generated after pruning; The results of our experiments are listed in Table3 . All the values are averaged over ten folds.... In PAGE 15: ... All the values are averaged over ten folds. We discovered an important fact from Table 2 and Table3 : for those datasets with more than 50 contextual rules after pruning, the average accuracy improvement of CLNB over NB is 6.3% (on 11 datasets), and the average accuracy improvement of CLNN over NN is 7.... ..."

### Table 1 Computational requirements

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### Table 5. Computational Requirements

### Table 5: Computational Requirements

### Table V. Computational requirements

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### Table 1 gives lower bounds on the value of N for which we expect to contain a permutation with probability one half. The rst column shows, to the nearest power of 2, the maximum value of m for which Pr(N; m) gt; 1 2 as computed from eq. (8). The second column shows N m as bounded by Lemma 3.1. Table 1 clearly indicates that N m is impractically large for m as small as 8, and is infeasible for m = 10. This infeasibility represents the basic limitation to all bit-by-bit methods. m Pr(N; m) gt; 0:5 N m 4

"... In PAGE 4: ... We prove (Lemma 3.1) that N m gt; e2m=m?m, which implies that for m gt; 8, approximately 269 functions must be selected to nd a permutation with probability one half (see Table1 in x3). Given that N was chosen su ciently large in Step 1 to ensure that contains a permutation, in Step 2 of the bit-by-bit method we must actually nd a permutation from .... In PAGE 8: ... This infeasibility represents the basic limitation to all bit-by-bit methods. m Pr(N; m) gt; 0:5 N m 4 24 0 5 27 22 6 213 27 7 224 216 8 245 234 9 280 269 10 2146 2133 Table1 : Size of N for which is expected to contain a permutation. 3.... In PAGE 14: ... In x3.3 it was shown that to guarantee the success of this algorithm to nd an arbitrary permutation, the size a must be large as compared to m (see Table1 in x4.... In PAGE 19: ...and were maximally nonlinear. We may also compare our results derived in Table1 of x4.0.... In PAGE 19: ...e may also compare our results derived in Table 1 of x4.0.2 with the results of Algorithm B. Table1 indicates that for m = 4, N = 16 functions are required to nd a permutation with probability one half. From Table 4 we see that whenever more than 16 functions in satisfying the restrictions for that run were found, over half the runs found a permutation.... In PAGE 19: ... From Table 4 we see that whenever more than 16 functions in satisfying the restrictions for that run were found, over half the runs found a permutation. Similarly, the bound from Table1 for m = 5 to yield a permutation with probability one half is N = 128. We nd that the results of Table 5 also agree with this bound.... In PAGE 19: ...Consider the case now where m = 6. From Table1 , we will require approximately 8000 functions before is expected to contain a 6-bit permutation. To estimate the size of the search tree that would be examined by the Algorithm B, we may use a Monte Carlo approximation technique [28].... In PAGE 19: ... We then infer that 6-bit permutations which satisfy the SAC and are maximally nonlinear can be found by Algorithm B using a large amount of computer resources. For m = 7, Table1 indicates that must have approximately 224 functions before a permutation is expected to be found. From the Monte Carlo technique, the number of nodes in the search tree to be examined by Algorithm B is greater than 240, and we conclude that Algorithm B is not feasible for m gt; 6.... ..."

### Table 6.4 RT0 space versus BDM1 space in terms of CPU time (sec) and computer storage (Kbytes) for a xed relative error: rel = kq ? qhk0; =kqk1; . Test case # 3, = :99. RT0 BDM1

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